Tag: matrices and determinants

Questions Related to matrices and determinants

If $\Delta =\begin{vmatrix} { a } _{ 11 } & { a } _{ 12 } & { a } _{ 13 } \ { a } _{ 21 } & { a } _{ 22 } & { a } _{ 23 } \ { a } _{ 31 } & { a } _{ 32 } & { a } _{ 33 } \end{vmatrix}$ and ${ A } _{ ij }$ is cofactors of ${ a } _{ ij }$, then the value of $\Delta $ is given by

  1. ${ a } _{ 11 }{ A } _{ 31 }+{ a } _{ 12 }{ A } _{ 32 }+{ a } _{ 13 }{ A } _{ 33 }$

  2. ${ a } _{ 11 }{ A } _{ 11 }+{ a } _{ 12 }{ A } _{ 21 }+{ a } _{ 13 }{ A } _{ 31 }$

  3. ${ a } _{ 21 }{ A } _{ 11 }+{ a } _{ 22 }{ A } _{ 12 }+{ a } _{ 23 }{ A } _{ 13 }$

  4. ${ a } _{ 11 }{ A } _{ 11 }+{ a } _{ 21 }{ A } _{ 21 }+{ a } _{ 31 }{ A } _{ 31 }$


Correct Option: D
Explanation:

$\Delta =\begin{vmatrix} { a } _{ 11 } & { a } _{ 12 } & { a } _{ 13 } \ { a } _{ 21 } & { a } _{ 22 } & { a } _{ 23 } \ { a } _{ 31 } & { a } _{ 32 } & { a } _{ 33 } \end{vmatrix}$


Also given ${ A } _{ ij }$ is cofactor of ${ a } _{ ij }$
$\Delta =a _{11} A _{11}+a _{21} A _{21}+a _{31} A _{31}$

$A=\left{\begin{array}{ll}
8 & 9\
10 & 11
\end{array}\right}$, then cofactor of $\mathrm{a} _{12}$ is:

  1. 11

  2. 10

  3. -11

  4. -10


Correct Option: D
Explanation:

By property of cofactor (2),
$A=\begin{bmatrix}
8 & 9\
10 & 11
\end{bmatrix}$
minor of $a _{12}=M _{12}=10$
So, cofactor of $a _{12}=M _{12}(-1)^{1+2}=-M _{12}=-10$

If $\triangle =\begin{bmatrix} { a } _{ 1 } & { b } _{ 1 } & { c } _{ 1 } \ { a } _{ 2 } & { b } _{ 2 } & { c } _{ 2 } \ { a } _{ 3 } & { b } _{ 3 } & { c } _{ 3 } \end{bmatrix}$ and ${A} _{2},{B} _{2},{C} _{2}$ are respectively cofactors of ${a} _{2},{b} _{2},{c} _{2}$ then ${a} _{1}{A} _{2}+{b} _{1}{B} _{2}+{c} _{1}{C} _{2}$ is equal to ?

  1. $-\triangle$

  2. $0$

  3. $\triangle$

  4. $none\ of\ these$


Correct Option: B
Explanation:

Co-factor of $ \displaystyle a _2 = (-1)^{i+j}\left|\begin{matrix} b _1 & c _1 \ b _3 & c _3 \end{matrix}\right| = - [b _1 \, c _3 - c _1 \, b _3] = A _2 $

i=2  j=1

of $ \displaystyle b _2 = (-1)^{2+2} \left|\begin{matrix} a _1 & c _1 \\ a _3 & c _3 \end{matrix}\right| = [a _1 \, c _3 - c _1 \, a _3] = B _2 $ 

of $ \displaystyle c _2 = (-1)^{2+3} \left|\begin{matrix} a _1 & b _1 \\ a _3 & b _3 \end{matrix}\right| = -[a _1 \, b _3 \, - b _1 \, a _3 ] = C _2 $

$ a _1 \, A _2 + b _1 \, B _2 +c _1 \, C _2 $

$ \displaystyle -a _1 \, b _1 \, c _3 + a _1 \, c _1 \, b _3 + b _1 \, a _1 \, c _3 - b _1 c _1 \, a _3 -c _1 \, a _1 \, b _3 + c _1 \, b _1 \, a _3 $

$ \displaystyle = 0 $

If $\Delta = \begin{vmatrix}a _1 & b _1 & c _1 \ a _2 & b _2 & c _2\ a _3 & b _3 & c _3\end{vmatrix}$ and $A _1, B _1, C _1$ denote the co-factors of $a _1, b _1, c _1$ respectively, then teh value os the determinant $\begin{vmatrix}A _1 & B _1 & C _1\ A _2 & B _2 & C _2\ A _3 & B _3 & C _3\end{vmatrix}$ is-

  1. $\Delta$

  2. $\Delta^2$

  3. $\Delta^3$

  4. $0$


Correct Option: B
Explanation:

Given,

$\Delta = \begin{vmatrix}a _1 & b _1 & c _1 \ a _2 & b _2 & c _2\ a _3 & b _3 & c _3\end{vmatrix}$ and $A _1, B _1, C _1$ denote the co-factors of $a _1, b _1, c _1$ respectively.
Now,
 $\begin{vmatrix}A _1 & B _1 & C _1\ A _2 & B _2 & C _2\ A _3 & B _3 & C _3\end{vmatrix}$
$=|adj \Delta|$
$=\Delta^{3-1}$
$=\Delta^2$.

If $\Delta  = \left| {\begin{array}{*{20}{c}}  {{a _1}}&{{b _1}}&{{c _1}} \   {{a _2}}&{{b _2}}&{{c _2}} \   {{a _3}}&{{b _3}}&{{c _3}} \end{array}} \right|$ and $A _2$, $B _2$, $C _2$ are respectively cofactors of $a _2,b _2,c _2$ then 


$a _1A _2+b _1B _2+c _1C _2$ is 

  1. $ - \Delta $

  2. $0$

  3. $\Delta $

  4. none of these


Correct Option: B
Explanation:
$\Delta =\begin{vmatrix} { a } _{ 1 } & { b } _{ 1 } & { c } _{ 1 } \\ { a } _{ 2 } & { b } _{ 2 } & { c } _{ 2 } \\ { a } _{ 3 } & { b } _{ 3 } & { c } _{ 3 } \end{vmatrix}$

$A _2=c _3\ b _1+b _1\ c _1$

$B _2=a _1\ c _3+a _3\ b _1$

$C _2=-a _1\ b _3+a _3\ b _1$

$\to \ a _1 A _2+b _1 B _2+c _1 C _2$

$a, b, c _3+a, c, b _3+b, a, c _3-b, a, c _3-b, c, a _3-c, a, b _3+b, c, a _3$

$=0$

$B$ is correct.

The value of a third order determinant is $11$, then the value of the square of the determinant formed by the cofactors will be?

  1. $11$

  2. $121$

  3. $1331$

  4. $14641$


Correct Option: D
Explanation:
third order determinant = determinant of $3\times 3$ matrix $A$
given $|A|=11$
det (cofactor matrix of $A$) =set (transpare of cofactor amtrix of $A$) (transpare does not change the det)
=det(adjacent of $A$)
$\left\{det\ (cofactor\ matrix\ of\ A) \right\}^2=\left\{det\ (adjacent\ of\ A)\right\}^2$
(Using for an $n\times n\ det\ (cofactor\ matrix\ of\ A)=det\ (A)^{n-1})$
we get
$det\ (cofactor\ matrix\ of\ A)^2=\left\{det (A)^{3-1}\right\}^2$
$=(11)^{2\times 2}=11^4$
$=146.41$

Consider the determinant, $\Delta=\begin{vmatrix} p & q & r \ x & y & z \ l & m & n \end{vmatrix}$ ${M} _{0}$ denotes the minor of an element in $i$th row and $j$th column and ${C} _{ij}$ denotes the cofactor of an element in $i$th row and $j$th column.
The value of $p.{C} _{21}+q.{C} _{22}+r.{C} _{23}$ is equal to

  1. $0$

  2. $-\Delta$

  3. $\Delta$

  4. ${\Delta}^{2}$


Correct Option: A
Explanation:

From the property of determinants, if any element is multiplied with cofactor of corresponding element of another row, and summed up for each element of original row, the sum comes out zero $\Rightarrow (A)$

The cofactor of the element $4$ in the determinant $\begin{vmatrix} 1 & 3 & 5 & 1\ 2 & 3 & 4 & 2\ 8 & 0 & 1 & 1\ 0 & 2 & 1 & 1\end{vmatrix}$ is?

  1. $4$

  2. $10$

  3. $-10$

  4. $-4$


Correct Option: A

If $A=\left[ \begin{matrix} { a } _{ 11 } & { a } _{ 12 } & { a } _{ 13 } \ { a } _{ 21 } & { a } _{ 22 } & { a } _{ 23 } \ { a } _{ 31 } & { a } _{ 32 } & { a } _{ 33 } \end{matrix} \right] $ and $C _{ij}$ is cofactor of $a _{ij}$ in $A$, then value of $|A|$ is given by

  1. $a _{11}C _{31}+a _{12}C _{32}+a _{13}C _{33}$

  2. $a _{11}C _{11}+a _{12}C _{21}+a _{13}C _{31}$

  3. $a _{21}C _{11}+a _{22}C _{21}+a _{23}C _{31}$

  4. $a _{11}C _{11}+a _{21}C _{21}+a _{31}C _{31}$


Correct Option: A

If $\begin{vmatrix} { a }^{ 2 }+{ \lambda  }^{ 2 } & ab+c\lambda  & ca-b\lambda  \ ab-c\lambda  & { b }^{ 2 }+{ \lambda  }^{ 2 } & bc+a\lambda  \ ca+b\lambda  & bc-a\lambda  & { c }^{ 2 }+{ \lambda  }^{ 2 } \end{vmatrix}\begin{vmatrix} \lambda  & c & -b \ -c & \lambda  & a \ b & -a & \lambda  \end{vmatrix}={ \left( 1+{ a }^{ 2 }+{ b }^{ 2 }+{ c }^{ 2 } \right)  }^{ 3 }$, then the value of $\lambda$ is

  1. 8

  2. 27

  3. 1

  4. -1


Correct Option: C
Explanation:

Let $D=\begin{vmatrix} \lambda  & c & -b \ -c & \lambda  & a \ b & -a & \lambda  \end{vmatrix}$


determinant of cofactors is

$D^{c}=\begin{vmatrix} { a }^{ 2 }+{ \lambda  }^{ 2 } & ab+c\lambda  & ca-b\lambda  \ ab-c\lambda  & { b }^{ 2 }+{ \lambda  }^{ 2 } & bc+a\lambda  \ ca+b\lambda  & bc-a\lambda  & { c }^{ 2 }+{ \lambda  }^{ 2 } \end{vmatrix}=D^2$

$\begin{vmatrix} { a }^{ 2 }+{ \lambda  }^{ 2 } & ab+c\lambda  & ca-b\lambda  \ ab-c\lambda  & { b }^{ 2 }+{ \lambda  }^{ 2 } & bc+a\lambda  \ ca+b\lambda  & bc-a\lambda  & { c }^{ 2 }+{ \lambda  }^{ 2 } \end{vmatrix}\begin{vmatrix} \lambda  & c & -b \ -c & \lambda  & a \ b & -a & \lambda  \end{vmatrix}={ \left( 1+{ a }^{ 2 }+{ b }^{ 2 }+{ c }^{ 2 } \right)  }^{ 3 }$

$\Rightarrow D^3= { \left( 1+{ a }^{ 2 }+{ b }^{ 2 }+{ c }^{ 2 } \right)  }^{ 3 }$ -------(1)
Now,
$D=\begin{vmatrix} \lambda  & c & -b \ -c & \lambda  & a \ b & -a & \lambda  \end{vmatrix}$

 $=\lambda(\lambda^2+a^2)-c(-\lambda c-ab)-b(ac-b\lambda)$
 $=\lambda(\lambda^2+a^2+b^2+c^2)$
from (1)
$\left(\lambda(\lambda^2+a^2+b^2+c^2)\right)^3={ \left( 1+{ a }^{ 2 }+{ b }^{ 2 }+{ c }^{ 2 } \right)  }^{ 3 }$
comparing on both sides gives
$\lambda^3=1$ and $\lambda^2=1$
$\therefore \lambda=1$
Hence, option C.